$\dfrac{ t - 4u }{ -9 } = \dfrac{ 8t - 4v }{ 6 }$ Solve for $t$.
Multiply both sides by the left denominator. $\dfrac{ t - 4u }{ -{9} } = \dfrac{ 8t - 4v }{ 6 }$ $-{9} \cdot \dfrac{ t - 4u }{ -{9} } = -{9} \cdot \dfrac{ 8t - 4v }{ 6 }$ $t - 4u = -{9} \cdot \dfrac { 8t - 4v }{ 6 }$ Multiply both sides by the right denominator. $t - 4u = -9 \cdot \dfrac{ 8t - 4v }{ {6} }$ ${6} \cdot \left( t - 4u \right) = {6} \cdot -9 \cdot \dfrac{ 8t - 4v }{ {6} }$ ${6} \cdot \left( t - 4u \right) = -9 \cdot \left( 8t - 4v \right)$ Distribute both sides ${6} \cdot \left( t - 4u \right) = -{9} \cdot \left( 8t - 4v \right)$ ${6}t - {24}u = -{72}t + {36}v$ Combine $t$ terms on the left. ${6t} - 24u = -{72t} + 36v$ ${78t} - 24u = 36v$ Move the $u$ term to the right. $78t - {24u} = 36v$ $78t = 36v + {24u}$ Isolate $t$ by dividing both sides by its coefficient. ${78}t = 36v + 24u$ $t = \dfrac{ 36v + 24u }{ {78} }$ All of these terms are divisible by $6$ $t = \dfrac{ {6}v + {4}u }{ {13} }$